Couple questions on work and conservation of energy

1. A woman at an airport is towing her 29.0 kg suitcase at constant speed by pulling on the strap at an angle above the horizontal. She pulls on the strap with a 40.0 N force, and the friction force on the suitcase is 12.0 N.

(a) What angle does the strap make with the horizontal?
(b) What normal force does the ground exert on the suitcase?

Ok I've figured out a and it is 72.54 degrees. Part b I submitted and it told me that the sign was wrong. I put in -9.2 N first and it said to change the sign then I tried 9.2 and it was "orders of magnitude" off. So on my final chance I'd like to get it correct this time. I could use help with this one. How do I figure out the normal force? I thought it was just the opposite of mg but someone else told me that was wrong. I used m-40sin72.54 but that was seemingly wrong. I haven't tried mg but would that be correct? What's the actual answer.

3. A crate of mass 12.5 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 17.5° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.15 m.

(a) How much work is done by gravity?
(b) How much mechanical energy is lost due to friction?
(c) How much work is done by the 100 N force?
(d) What is the change in kinetic energy of the crate?
(e) What is the speed of the crate after being pulled 5.15 m?

For this one I have part a and b correct. a being -189.7 J and b being 240.7 J. Part C it tells me I was within 10% of the correct value and I put 491.2 J. Part D I was between 10 and 100% away from the correct answer with 60.8 J. Part E I was between 10 and 100% away. I've submitted those answers once and only part a and b are correct as I've said. I've tried to do it differently but I can't figure it out.

6. Tarzan swings on a 34.0 m long vine initially inclined at an angle of 40.0° with the vertical.
(a) What is his speed at the bottom of the swing if he starts from rest?
(b) What is his speed at the bottom of the swing if he pushes off with a speed of 6.00 m/s?

Ok I've calculated a to be 12.5 m/s and b to be 13.9 m/s. I have yet to submit this answer because I only have one chance left. I submitted different answers by doing it a different way that I think is incorrect but it was within 10-100%.

The way I did it was to find the vertical distance when Tarzan is at the vertical through 34sin50. Then I subtracted 34m-26.046. So then I had the height that Tarzan starts above when he is at the halfway point at the vertical. So I went through and used the conservation of energy for a. mgh=1/2 mv^2 where the masses aren't needed and came up with my answer for a which was 12.5.

For part b I used mgh+1/2 mv^2 initial=1/2mv^2 final. I then plugged in all the values and calculated the velocity to be 13.9 m/s.

Just tell me if those answers are correct so I can submit them. For the rest of them I could use some help for getting the correct answers.

Anyway, my name is Geoff and I'd appreciate any help you can give. Thanks.

1b The suitcase is not accelerating in the y-direction, so the sum of all the y-components of all of the forces should be zero. That implies that the weight of the suitcase should be balanced by the normal force and the y-component of the pull.

3 The work done by the 100 N force (a nonconservative force) [itex]W_{NC}[/itex]
determines the change in the mechanical energy of the crate
[tex]W_{NC}=\Delta KE + \Delta PE[/tex]
Remember that the change in potential energy of the crate is given by the negative of the work done by gravity on it.